摘取于http://blog.csdn.net/kenden23/article/details/37519883: 找到规律之后本题就是水题了,不过找规律也不太容易的,证明这个规律成立更加不容易. 本题就是求step和mod如果GCD(最大公约数位1)那么就是Good Choice,否则为Bad Choice 为什么这个结论成立呢? 因为当GCD(step, mod) == 1的时候,那么第一次得到序列:x0, x0 + step, x0 + step…… 那么mod之后,必然下一次重复出现比…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1014 Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33120    Accepted Submission(s): 13137 Problem Description Computer simulati…

Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29336    Accepted Submission(s): 11694 Problem Description Computer simulations often require random numbers. One way to generat…

Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24381    Accepted Submission(s): 9642 Problem Description Computer simulations often require random numbers. One way to generat…

Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32990    Accepted Submission(s): 13081 Problem Description Computer simulations often require random numbers. One way to generat…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1014 题目意思:给出 STEP 和 MOD,然后根据这个公式:seed(x+1) = [seed(x) + STEP] % MOD,问是否在一个周期里可以产生 0 - mod-1 的数.可以的话输出 "Good Choice", 否则输出 "Bad Choice". 好久以前留下来的问题了,以前觉得题目意思又长,以为是很难的题目......今天看<短码之美>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1014 解题思路: 1. 把题目意思读懂后,明白会输入两个数,然后根据题中的公式产生一系列伪随机数,看这些数是不是能够包含0~MOD-1.如果产生不了就输出“Good Choice”,否则输出“Bad Choice”. 2. 全部假使x从0开始,设STEP为a,MOD为b.如果说a,b存在倍数关系,即假设最小存在2倍关系,那么就会有b=2a.x初始为0,第一步之后为a,第二步之后为0,之后a.0交替出…

找到规律之后本题就是水题了.只是找规律也不太easy的.证明这个规律成立更加不easy. 本题就是求step和mod假设GCD(最大公约数位1)那么就是Good Choice,否则为Bad Choice 为什么这个结论成立呢? 由于当GCD(step, mod) == 1的时候.那么第一次得到序列:x0, x0 + step, x0 + step-- 那么mod之后,必定下一次反复出现比x0大的数必定是x0+1,为什么呢? 由于(x0 + n*step) % mod. 且不须要考虑x0 % mo…

#include<iostream> #include <cstdio> #include <cstring> using namespace std; int main(void) { int s,m,i; /*int seed=0; int a[10005]={0}; //memset(a,0,sizeof(a));*/ while(scanf("%d%d",&s,&m)!=EOF); { int seed=0; int a[10…

http://acm.hdu.edu.cn/showproblem.php?pid=1014 题目的英文实在是太多了 ,搞不懂. 最后才知道是用公式seed(x+1) = [seed(x) + STEP] % MOD 来计算随机数 ,问是否满足随机数. 初级版本: 思路:把所有的用该公式计算出来的数(存在数组中)都遍历出来,然后排序.由于数字是在0 到mod-1 之间,所以数组的下标必然等于数组的值,有一个不等于,就是bad chioce #include <stdio.h> #include…

Problem Description Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form seed(x+1) = [seed(x) + STEP] % MOD where '%' is the modulus operator. Such a function will generate pseudo-…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28661    Accepted Submission(s): 11402 Problem Description Computer simulations often require random numbers. One way to generate pseudo-random nu…

Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21176 Accepted Submission(s): 8294 Problem Description Computer simulations often require random numbers. One way to generate pseu…

Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35190 Accepted Submission(s): 14002 Problem Description Computer simulations often require random numbers. One way to generate pseu…

Uniform Generator  Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form where ``  " is the modulus operator. Such a function will generate pseudo-random numbers (seed) between 0 an…

HDU 1005 Number Sequence(数论) Problem Description: A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multipl…

Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26820 Accepted Submission(s): 10629 Problem Description Computer simulations often require random numbers. One way to generate pse…

Least Common Multiple (HDU - 1019) [简单数论][LCM][欧几里得辗转相除法] 标签: 入门讲座题解 数论 题目描述 The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7…

七夕节 (HDU - 1215) [简单数论][找因数] 标签: 入门讲座题解 数论 题目描述 七夕节那天,月老来到数字王国,他在城门上贴了一张告示,并且和数字王国的人们说:"你们想知道你们的另一半是谁吗?那就按照告示上的方法去找吧!" 人们纷纷来到告示前,都想知道谁才是自己的另一半.告示如下: 数字N的因子就是所有比N小又能被N整除的所有正整数,如12的因子有1,2,3,4,6. 你想知道你的另一半吗? Input 输入数据的第一行是一个数字T(1<=T<=500000)…

http://acm.hdu.edu.cn/showproblem.php?pid=1014 给出式子seed(x+1) = [seed(x) + STEP] % MOD seed初始为0,给出STEP和MOD的值 问seed能否取到0~(MOD - 1)之间的所有值 简单模拟 # include <stdio.h> int main() { int Step, Mod, i, Seed, Flag[100005]; while(scanf("%d %d",&Ste…

H - Visible Trees Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2841 Description There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) poi…

http://acm.hdu.edu.cn/showproblem.php?pid=4036 一开始以为需要用斜抛,结果发现只需要用能量守恒定律?+与最大速度的坏土豆速度保持一致 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int maxn=1e3+3; int n,m; double w; doub…

本文参考资料:http://hi.baidu.com/bnjyjncwbdbjnzr/item/1f997cfdd225d5d143c36a58 题意:一个生成随机数的函数, Seed[x+1] = ( seed[x] + STEP ) % MOD 输入step和mod,问能否生成0~MOD-1之间所有的数,是Good Choice,否则Bad Choice 题意其实就是:给出S和M,求0*S%M,1*S%M,2*S%M......(M-1)*S%M能否组成一个集合包含0.1....M-1:(这…

http://acm.hdu.edu.cn/showproblem.php? pid=1124 題目好長好長,好可怕,看完腎都萎了,以後肯定活不長.我可不能死在這種小事上,小灰灰我勵志死在少女的超短裙下~~~哈哈,所以我就猥瑣的叫 旁邊的小師妹幫我翻譯了,我是不是非常禽獸,嘻嘻~~~ 題目大意呢,就是給一個數,要你求出它的階乘的得到的結果後面有幾個0. 解析: 一看就是簡單數論啦.跟數因子有關.最小素因子并且相乘能得到10的(就是後面有0的)就是2*5啦.因為一個數的階乘2的因子明顯比5的因子要…

继续数论.. Problem Description WhereIsHeroFrom:            Zty,what are you doing ? Zty:                                    Iwant to calculate N!...... WhereIsHeroFrom:            Soeasy! How big N is ? Zty:                                    1<=N <=100…

题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1852 这道题和HDU1452类似. 题意:给你一个n.k,让你求2008^n所有因子的和(包括1和本身)%k,得到m,然后输出2008^m%k. 题解:看我HDU1452题,这里有一点需要注意的是: s=(2^(3n+1)-1)(251^(n+1)-1)/250 因为gcd(250,k)不一定等于1,所以不能用求逆元的方法求解, 而k很小,所以我们可以将k乘以250,然后在进行,最后结果一定可以整…

题目链接:hdu 4861 Couple doubi 题目大意:两个人进行游戏,桌上有k个球,第i个球的值为1i+2i+⋯+(p−1)i%p,两个人轮流取,假设DouBiNan的值大的话就输出YES,否则输出NO. 解题思路: 首先是DouBiNan先取,所以肯定优先选取剩余中值最大的,于是不存在说DouBiNan值小的情况,仅仅有大于和小于. 然后,对于val(i)=1i+2i+⋯+(p−1)i%p来说,仅仅有当i=ϕ(p)=p−1(p为素数)时,val(i)=p−1,其它情况下val(i)=…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6169 题意:给了区间L,R,求[L,R]区间所有满足其最小质数因子为k的数的和. 解法: 我看了这篇blog.http://blog.csdn.net/wubaizhe/article/details/77484454#cpp 先放在这里,明天来补推导过程. #include <bits/stdc++.h> using namespace std; typedef long long LL; co…

传送门 参考资料: [1]:https://www.2cto.com/kf/201308/233613.html 题意,题解在上述参考资料中已经介绍的非常详细了,接下来的内容只是记录一下我的理解: 我的学习记录: 定义 f(x) : x的因子个数: φ(x) : x之前与x互素的数的个数: 那么 F(x) = x - f(x) - φ(x) + 1; 为什么要 +1 呢? 因为 f(x) 和 φ(x) 同时包含 1 这个数,所以要加上多减去的 1: 根据算术基本定理: 任何一个大于1的自然数 N…

Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form seed(x+1) = [seed(x) + STEP] % MOD where '%' is the modulus operator. Such a function will generate pseudo-random numbers (seed…